# Determinant of surjective endomorphism of $R^{\oplus n}$

Let $$R$$ be a commutative ring with identity and let $$f\colon R^{\oplus n}\to R^{\oplus n}$$ be a surjective linear mapping. Show that $$\det f\in R^\times$$.

This question was inspired by A surjective homomorphism between finite free modules of the same rank .

I am curious about what is the proof mentioned in the question .

Call $$e^1,\cdots, e^n$$ the canonical basis of $$R^n$$, call $$F$$ the matrix that represents $$f$$ in that basis and call $$F^1,\cdots, F^n$$ the colums of $$F$$. Saying that $$\operatorname{col}F=R^n$$ means that there are $$a_{ij}$$ such that for all $$j$$, $$\sum_{i=1}^n a_{ij}F^i=e^j$$. In other words, that there is an $$n\times n$$ matrix $$A$$ such that $$FA=I$$. Now, by Binet's theorem, $$\det(F)\det(A)=\det(FA)=\det I=1$$, q.e.d.
Why Binet holds in commutative rings with 1: Consider the polynomial ring $$R[X_1,\cdots, X_n]$$. There is a natural homomorphism $$u:\Bbb Z[X_1,\cdots, X_n]\to R[X_1,\cdots, X_n]$$, $$u_R(\sum_{J\in\Bbb N^n} a_JX^J)=\sum_{J\in\Bbb N^n}(a_J)_RX^J$$, where $$n_R:=\underbrace{1+\cdots+1}_{n\text{ times}}$$. Also, given a point $$p\in R^n$$, call $$\nu_p:R[X_1,\cdots,X_n]\to R$$ the map $$\nu_p(f)=f(p)$$. Now, Binet for matrices in $$R^{n\times n}$$ means that given the following polynomial $$b\in \Bbb Z[X_{ij},Y_{ij}\,:\, 1\le i,j\le n]$$ $$b=\left(\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n\sum_{j=1}^nX_{ij}Y_{j,\sigma(i)}\right)-\left(\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^nX_{i,\sigma(i)}\right)\left(\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^nY_{i,\sigma(i)}\right)$$
we have $$\nu_p(u_R(b))=0$$ for all $$p\in R^{n\times n}\times R^{n\times n}$$. However, by the identity principle of polynomials, proving Binet in the special case where $$R$$ is any infinite domain (for instance, the field $$\Bbb R$$ or $$\Bbb Q$$) implies that $$b=0$$.
• Another approach to prove that $\det AB=\det A\det B$ for non-fields is using $\mathsf{\Lambda}_R^n$ functor. Or just using the free ring of rank $2n^2$ and its fractions field Oct 7 at 6:19